$\sum\limits_{n=1}^{\infty } \dfrac{4(x-5)^n}{n^2\cdot(-4)^n}$ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $1 \le x \le 9$ (Choice B) B $1 < x \le 9$ (Choice C) C $0 < x \le 8$ (Choice D) D $0\le x \le 8$
Explanation: We use the ratio test. For $x\neq5$, let $a_n=\dfrac{4(x-5)^n}{n^2\cdot(-4)^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x-5}{4}\right| $ The series converges when $\left| \dfrac{x-5}{4}\right| <1$, which is when $1<x<9 $. Now let's check the endpoints, $x=1$ and $x=9$. Letting $x=1$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{4(1-5)^n}{n^2\cdot(-4)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{4(-4)^n}{n^2\cdot(-4)^n} \\\\ &=4\sum\limits_{n=1}^{\infty } \dfrac{1}{n^2} \end{aligned}$ By the $p$ -series test, we know this series converges. Letting $x=9$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{4(9-5)^n}{n^2\cdot(-4)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{4(4)^n}{n^2\cdot(-4)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{4(-1)^n(-4)^n}{n^2\cdot(-4)^n} \\\\ &=\sum\limits_{n=1}^{\infty } (-1)^n\cdot \dfrac{4}{n^2} \end{aligned}$ By the alternating series test, we know this series converges. In conclusion, the interval of convergence is $1 \le x \le 9$.